/_ A = (2x+4)°, /_B = (y+3)° , /_ C= (2y+10)°,/_D=(4x-5)°

Now, /_A+ /_C = 180° ( opp angles of a cyclic quad are supplementary ).......1

and /_B+/_D = 180 ( same as above)................2

putting the values in 1 and 2,(2x+4)° + (2y+10)° = 180°

Slly, (y+3)° +(4x-5)°= 180°.= (4x + y - 2)= 180.

multiplying eq. 1 by 2, 2 X {(2x+4)° + (2y+10)° = 180°}

= 2X ( 2x+2y+14)=4x + 4y +28.=360°

Finally, subtracting 2 from 1

,(4x + 4y +28.=360°) - (4x + y - 2)=180.

= 3y + 30 = 360-180= 180.

3y = 180-30=150.

y = 150/3 = 50°.

putting value of y in eq 2.

(4x + 50 - 2)= 180.

we get x = 180-48/4 = 33°.

Hence, /_A= 2x33+4= 70°/_ B = 50+3= 53°/_C= 180 - 70 = 110°/_D= 180-53 = 127°